Calculus Examples

$f (x) = \sqrt{\sin (π (x - 1))} + \sqrt{4 - x^{2}}$

Step 1

Set the radicand in $\sqrt{\sin (π (x - 1))}$ greater than or equal to $0$ to find where the expression is defined.

$\sin (π (x - 1)) \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$π (x - 1) \geq \arcsin (0)$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $π (x - 1)$ .

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Step 2.2.1.1

Simplify by multiplying through.

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Step 2.2.1.1.1

Apply the distributive property.

$π x + π \cdot - 1 \geq \arcsin (0)$

Step 2.2.1.1.2

Move $- 1$ to the left of $π$ .

$π x - 1 \cdot π \geq \arcsin (0)$

Step 2.2.1.2

Rewrite $- 1 π$ as $- π$ .

$π x - π \geq \arcsin (0)$

Step 2.3

Simplify the right side.

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Step 2.3.1

The exact value of $\arcsin (0)$ is $0$ .

$π x - π \geq 0$

Step 2.4

Add $π$ to both sides of the inequality.

$π x \geq π$

Step 2.5

Divide each term in $π x \geq π$ by $π$ and simplify.

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Step 2.5.1

Divide each term in $π x \geq π$ by $π$ .

$\frac{π x}{π} \geq \frac{π}{π}$

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Cancel the common factor of $π$ .

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Step 2.5.2.1.1

Cancel the common factor.

$\frac{π x}{π} \geq \frac{π}{π}$

Step 2.5.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{π}{π}$

Step 2.5.3

Simplify the right side.

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Step 2.5.3.1

Cancel the common factor of $π$ .

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Step 2.5.3.1.1

Cancel the common factor.

$x \geq \frac{π}{π}$

Step 2.5.3.1.2

Rewrite the expression.

$x \geq 1$

Step 2.6

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$π x - π = π - 0$

Step 2.7

Solve for $x$ .

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Step 2.7.1

Subtract $0$ from $π$ .

$π x - π = π$

Step 2.7.2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.7.2.1

Add $π$ to both sides of the equation.

$π x = π + π$

Step 2.7.2.2

Add $π$ and $π$ .

$π x = 2 π$

Step 2.7.3

Divide each term in $π x = 2 π$ by $π$ and simplify.

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Step 2.7.3.1

Divide each term in $π x = 2 π$ by $π$ .

$\frac{π x}{π} = \frac{2 π}{π}$

Step 2.7.3.2

Simplify the left side.

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Step 2.7.3.2.1

Cancel the common factor of $π$ .

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Step 2.7.3.2.1.1

Cancel the common factor.

$\frac{π x}{π} = \frac{2 π}{π}$

Step 2.7.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{2 π}{π}$

Step 2.7.3.3

Simplify the right side.

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Step 2.7.3.3.1

Cancel the common factor of $π$ .

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Step 2.7.3.3.1.1

Cancel the common factor.

$x = \frac{2 π}{π}$

Step 2.7.3.3.1.2

Divide $2$ by $1$ .

$x = 2$

Step 2.8

Find the period of $\sin (π (x - 1))$ .

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Step 2.8.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 2.8.2

Replace $b$ with $π$ in the formula for period.

$\frac{2 π}{| π |}$

Step 2.8.3

$π$ is approximately $3.14159265$ which is positive so remove the absolute value

$\frac{2 π}{π}$

Step 2.8.4

Cancel the common factor of $π$ .

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Step 2.8.4.1

Cancel the common factor.

$\frac{2 π}{π}$

Step 2.8.4.2

Divide $2$ by $1$ .

$2$

Step 2.9

The period of the $\sin (π (x - 1))$ function is $2$ so values will repeat every $2$ radians in both directions.

$x = 1 + 2 n, 2 + 2 n$ , for any integer $n$

Step 2.10

Consolidate the answers.

$x = 1 n$ , for any integer $n$

Step 2.11

Use each root to create test intervals.

$0 < x < 1$

$1 < x < 2$

Step 2.12

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.12.1

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 2.12.1.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 2.12.1.2

Replace $x$ with $0.5$ in the original inequality.

$\sin (π ((0.5) - 1)) \geq 0$

Step 2.12.1.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.12.2

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

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Step 2.12.2.1

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Step 2.12.2.2

Replace $x$ with $1.5$ in the original inequality.

$\sin (π ((1.5) - 1)) \geq 0$

Step 2.12.2.3

The left side $1$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.12.3

Compare the intervals to determine which ones satisfy the original inequality.

$0 < x < 1$ False

$1 < x < 2$ True

$0 < x < 1$ False

$1 < x < 2$ True

Step 2.13

The solution consists of all of the true intervals.

$1 + 2 n \leq x \leq 2 + 2 n$ , for any integer $n$

Step 3

Set the radicand in $\sqrt{4 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$4 - x^{2} \geq 0$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $4$ from both sides of the inequality.

$- x^{2} \geq - 4$

Step 4.2

Divide each term in $- x^{2} \geq - 4$ by $- 1$ and simplify.

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Step 4.2.1

Divide each term in $- x^{2} \geq - 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 4}{- 1}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 4}{- 1}$

Step 4.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 4}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $- 4$ by $- 1$ .

$x^{2} \leq 4$

Step 4.3

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} \leq \sqrt{4}$

Step 4.4

Simplify the equation.

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Step 4.4.1

Simplify the left side.

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Step 4.4.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{4}$

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Simplify $\sqrt{4}$ .

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Step 4.4.2.1.1

Rewrite $4$ as $2^{2}$ .

$| x | \leq \sqrt{2^{2}}$

Step 4.4.2.1.2

Pull terms out from under the radical.

$| x | \leq | 2 |$

Step 4.4.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| x | \leq 2$

Step 4.5

Write $| x | \leq 2$ as a piecewise.

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Step 4.5.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 4.5.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 2$

Step 4.5.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 4.5.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 2$

Step 4.5.5

Write as a piecewise.

${\begin{cases} x \leq 2 & x \geq 0 \\ - x \leq 2 & x < 0 \end{cases}$

Step 4.6

Find the intersection of $x \leq 2$ and $x \geq 0$ .

$0 \leq x \leq 2$

Step 4.7

Solve $- x \leq 2$ when $x < 0$ .

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Step 4.7.1

Divide each term in $- x \leq 2$ by $- 1$ and simplify.

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Step 4.7.1.1

Divide each term in $- x \leq 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{2}{- 1}$

Step 4.7.1.2

Simplify the left side.

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Step 4.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{2}{- 1}$

Step 4.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{2}{- 1}$

Step 4.7.1.3

Simplify the right side.

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Step 4.7.1.3.1

Divide $2$ by $- 1$ .

$x \geq - 2$

Step 4.7.2

Find the intersection of $x \geq - 2$ and $x < 0$ .

$- 2 \leq x < 0$

Step 4.8

Find the union of the solutions.

$- 2 \leq x \leq 2$

Step 5

The domain is all values of $x$ that make the expression defined.

Set-Builder Notation:

${x | - 2 \leq x \leq 2, x = 1 + 2 n, x = 2 + 2 n}$

Step 6